Find the group of permutations on $\{1, 2, 3, 4\}$ which leaves the symmetric polynomial invariant

Find the group of permutations on $\{1, 2, 3, 4\}$ which leaves the symmetric polynomial $x_1 x_2+x_3x_4$ invariant.

A polynomial $f(x_1, . . . , x_n)$ is invariant under $S_n$ if for all $\pi \in S_n$ $$f(\pi(x_1), . . . , \pi(x_n)) = f(x_1, . . . , x_n)$$ But here how will I find the permutation such that the polynomial is invariant.

• The 2-cycle $(1,2)$ is in the group. Why? – Akiva Weinberger Jan 13 '17 at 2:14
• Where did you get the question from? $x_1x_2+ x_3x_4$ is not a symmetric polynomial. – Rob Arthan Jan 13 '17 at 2:22
• It is a symmetric polynomial. – Sachchidanand Prasad Jan 13 '17 at 2:24
• By definition a symmetric polynomial is a polynomial that is invariant under permutations of its variables. If $x_1x_2 + x_3x_4$ were a symmetric polynomial, then the answer to your question would be trivial. – Rob Arthan Jan 13 '17 at 2:31

Well, if $\sigma\in G$, the group you want to find, then you have: $$\sigma(x_1)\sigma(x_2)+\sigma(x_3)\sigma(x_4) = x_1x_2+x_3x_4$$.

The key here is that the above identity should be thought of as an equality between two polynomials. Can you take it from here?

• Then we can say that $\sigma$ is the identity element also. Except identity is there any? – Sachchidanand Prasad Jan 13 '17 at 2:17
• no, there are more. look at $\sigma = (x_2,x_1, x_3, x_4)$, for example. – dezdichado Jan 13 '17 at 2:19
• Even $(1 2), (3,4)$ this will also work. – Sachchidanand Prasad Jan 13 '17 at 2:19
• yes, but you are looking for all such $\sigma$s that make it work. do you understand the question itself actually? – dezdichado Jan 13 '17 at 2:21
• The permutation group acts on the indices not on the values of the polynomial: $$\sigma(x_1)\sigma(x_2)+\sigma(x_3)\sigma(x_4) = x_1x_2+x_3x_4$$ should read: $$x_{\sigma(1)}x_{\sigma(2)}+x_{\sigma(3)}x_{\sigma(4)} = x_1x_2+x_3x_4$$ – Rob Arthan Jan 13 '17 at 2:40